is merely a product of three consecutive integers, and we know that exactly one integer will be a multiple of 3, so we know that this expression will have a factor of 3, and so is divisible by 3.

Second Part

for and

Now as 99a and c(c+1)(c-1) are multiples of 3, therefore b(10-b) must also be a multiple of 3, so we require b = {1, 3, 4, 6, 7, 9}.
We also can see than c(c+1)(c-1) must be greater than or equal to 99, which implies that must be greater than 99.
This implies that c = {5, 6, 7, 8, 9}.

By rewriting the equation:

We can write down all possible values of the LHS, as we just look for the multiples of 99, with respect to the range of a:
99a = {99, 198, 297, 396, 495, 594, 693, 792, 891}

We can also write down all possible values of c(c+1)(c-1), with respect to our possible values of c:

We can also write down all the possible values of b(10-b) with respect to our set of b:
b(10-b) = {9, 21, 24}

Notice that, going back to the equation 99a = c(c+1)(c-1) - b(10-b), the maximum of the RHS is 720 - 9 = 711, so we can reject the two higher values of 99a, ie: 99a = {99, 198, 297, 396, 495, 594, 693}

The area of the rectangle ABCD is pq. It's centre of mass is at the point

The ara of the triangle ABX is qr/2 and it's centre of mass is

Therefore we have:

Equating the coordinates, we get:

And also:

Second Part

If , then r would be greater than q, which causes X not to lie in the closed interval [B,C], so the area cut away from the lamina would not be a triangle, however we are told it is a triangle, therefore r < q.